Problem 3.102

# Problem 3.102 - F σ π D ⋅ σ ⋅ cos θ ⋅ = π D ⋅...

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Problem *3.102 [Difficulty: 3] Given: Glass hydrometer used to measure SG of liquids. Stem has diameter D=5 mm, distance between marks on stem is d=2 mm per 0.1 SG. Hydrometer floats in kerosene (Assume zero contact angle between glass and kerosene). Find: Magnitude of error introduced by surface tension. Solution: We will apply the hydrostatics equations to this system. Governing Equations: F buoy ρ g V d = (Buoyant force is equal to weight of displaced fluid) d = 2 mm/0.1 SG F B y D = 5 mm Kerosene F σ Assumptions: (1) Static fluid (2) Incompressible fluid (3) Zero contact angle between ethyl alcohol and glass The surface tension will cause the hydrometer to sink h lower into the liquid. Thus for this change: Σ F z F buoy F σ = 0 = The change in buoyant force is: F buoy ρ g V = ρ g π 4 D 2 h = The force due to surface tension is:
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Unformatted text preview: F σ π D ⋅ σ ⋅ cos θ ( ) ⋅ = π D ⋅ σ ⋅ = Thus, ρ g ⋅ π 4 ⋅ D 2 ⋅ ∆ h ⋅ π D ⋅ σ ⋅ = Upon simplification: ρ g ⋅ D ⋅ ∆ h ⋅ 4 σ = Solving for ∆ h: ∆ h 4 σ ⋅ ρ g ⋅ D ⋅ = From Table A.2, SG = 1.43 and from Table A.4, σ = 26.8 mN/m Therefore, ∆ h 4 26.8 × 10 3 − × N m ⋅ m 3 1430 kg ⋅ × s 2 9.81 m ⋅ × 1 5 10 3 − × m ⋅ × kg m ⋅ s 2 N ⋅ × = ∆ h 1.53 10 3 − × m = So the change in specific gravity will be: ∆ SG 1.53 10 3 − × m ⋅ 0.1 2 10 3 − × m ⋅ × = ∆ SG 0.0765 = From the diagram, surface tension acts to cause the hydrometer to float lower in the liquid. Therefore, surface tension results in an indicated specific gravity smaller than the actual specific gravity....
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## This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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